传送门:牛客 - 练习赛67F
思路分析
两个区间内的直径有传递关系,可以通过两组端点两两组合得到
AC代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 1e6+10;
#define int long long
vector<pair<int,int>>e[N];
int fa[N],siz[N],d[N],son[N],top[N],dep[N];
void dfs1(int u,int f) {
siz[u]=1;
fa[u]=f;
dep[u]=dep[f]+1;
for(auto x:e[u]) {
int v=x.fi;
int w=x.se;
if(v==f) continue;
d[v]=d[u]+w;
dfs1(v,u);
siz[u]+=siz[v];
if(siz[v]>siz[son[u]]) son[u] = v;
}
}
void dfs2(int u,int t) {
top[u]=t;
if(!son[u]) return ;
dfs2(son[u],t);
for(auto x:e[u]) {
int v=x.fi;
int w=x.se;
if(v==fa[u] || v==son[u]) continue;
dfs2(v,v);
}
}
int lca(int x,int y) {
while(top[x]!=top[y]) {
if(dep[top[x]]<dep[top[y]]) swap(x,y);
x=fa[top[x]];
}
return dep[x]>dep[y]?y:x;
}
int dis(int x,int y) {
return d[x]+d[y]-2*d[lca(x,y)];
}
struct node {
int l,r,x,y,w;
} tree[N<<2];
#define ls x<<1
#define rs x<<1|1
#define mid (tree[x].l+tree[x].r>>1)
node merge(node a,node b) {
node ans = (a.w>b.w?a:b);
int tmp = dis(a.x,b.x);
if(tmp>ans.w) {
ans.w=tmp;
ans.x=a.x;
ans.y=b.x;
}
tmp = dis(a.x,b.y);
if(tmp>ans.w) {
ans.w=tmp;
ans.x=a.x;
ans.y=b.y;
}
tmp = dis(a.y,b.x);
if(tmp>ans.w) {
ans.w=tmp;
ans.x=a.y;
ans.y=b.x;
}
tmp = dis(a.y,b.y);
if(tmp>ans.w) {
ans.w=tmp;
ans.x=a.y;
ans.y=b.y;
}
return ans;
}
void build(int x,int l,int r) {
tree[x]= {l,r,0,0,0};
if(l==r) {
tree[x].x=tree[x].y=l;
return ;
}
build(ls,l,mid);
build(rs,mid+1,r);
tree[x] = merge(tree[ls],tree[rs]);
tree[x].l=l;
tree[x].r=r;
}
node query(int x,int l,int r) {
if(l<=tree[x].l && tree[x].r<=r) {
return tree[x];
}
if(r<=mid) return query(ls,l,r);
else if(l>mid) return query(rs,l,r);
else return merge(query(ls,l,r),query(rs,l,r));
}
signed main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
int n,q;
read(n,q);
for(int i=1;i<n;i++){
int u,v,w;
read(u,v,w);
e[u].pb(mp(v,w));
e[v].pb(mp(u,w));
}
dfs1(1,0);
dfs2(1,1);
build(1,1,n);
while(q--){
int l,r;
read(l,r);
printf("%lld\n",query(1,l,r).w);
}
return 0;
}